Find $\displaystyle lim_{x\to\infty} x[(1+\frac{1}{x})^x-e]$ [Suggestion: Let $\displaystyle u=\frac{1}{x}$, and take the limit as $\displaystyle u\to0$.]
Setting $\displaystyle x=\frac{1}{u}$ and applying several times l'Hopital's rule is ...
$\displaystyle \displaystyle \lim_{u \rightarrow 0} \frac{(1+u)^{\frac{1}{u}} - e}{u}= \lim_{n \rightarrow 0} \frac{(1+u)^{\frac{1}{u}}}{1+u}\ \frac {u- \ln (1+u)}{u^{2}} = e\ \lim_{u \rightarrow 0} \frac{1- \frac{1}{1+u}}{2u} = $
$\displaystyle \displaystyle = e \lim_{u \rightarrow 0} \frac{1}{2\ (1+u)^{2}} = \frac{e}{2}$
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$\displaystyle \chi$ $\displaystyle \sigma$